Compact Polish Space

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Tolaso J Kos
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Compact Polish Space

#1

Post by Tolaso J Kos »

Let \( X = [0,+\infty)\cup\{+\infty\} \). Endow it with the metric

$$\rho(x, y)=|\arctan x - \arctan y |$$

Prove that \( X \) under \( \rho \) is separable, complete and compact.
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Papapetros Vaggelis
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Re: Compact Polish Space

#2

Post by Papapetros Vaggelis »

We define \(\displaystyle{\arctan\,(+\infty)=\dfrac{\pi}{2}}\) .

1. The function \(\displaystyle{\rho}\) is a metric function on \(\displaystyle{X}\) :

Obviously, \(\displaystyle{\rho(x,y)\in\mathbb{R}\cap \left[0,+\infty\right)}\) and

\(\displaystyle{\rho(x,y)=0\iff x=y}\) . Also, \(\displaystyle{\rho(x,y)=\rho(y,x)\,,\forall\,x\,,y\in X}\) .

Let \(\displaystyle{x\,,y\,,z\in X}\) . Then,

\(\displaystyle{\rho(x,y)=\left|\arctan\,x-\arctan\,y\right|\leq \left|\arctan\,x-\arctan\,z\right|+\left|\arctan\,z-\arctan\,y\right|=\rho(x,z)+\rho(z,y)}\) .

2. We define \(\displaystyle{f:\left(X,\rho\right)\longrightarrow \left(\left[0,\dfrac{\pi}{2}\right],\left|\cdot\right|\right)}\)

by \(\displaystyle{f(x)=\arctan\,x}\) . The function \(\displaystyle{f}\) is well defined, one to one to \(\displaystyle{X}\)

and onto \(\displaystyle{\left[0,\dfrac{\pi}{2}\right]}\) . Also,

\(\displaystyle{\left|f(x)-f(y)\right|=\left|\arctan\,x-\arctan\,y\right|=\rho(x,y)\,,\forall\,x\,,y\in X}\), which means that

the function \(\displaystyle{f}\) is an isometry and additionally,

\(\displaystyle{\rho(f^{-1}(x),f^{-1}(y))=\left|x-y\right|\,,\forall\,x\,,y\in\left[0,\dfrac{\pi}{2}\right]}\) .

Therefore, the functions \(\displaystyle{f\,,f^{-1}}\) are continuous (take \(\displaystyle{\delta(\epsilon)=\epsilon\,,\epsilon>0}\))

and the metric spaces \(\displaystyle{\left(X,\rho\right)\,\,,\left(\left[0,\dfrac{\pi}{2}\right],\left|\cdot\right|\right)}\)

are homeomorphic. The metric space \(\displaystyle{\left(\left[0,\dfrac{\pi}{2}\right],\left|\cdot\right|\right)}\) is complete

as a closed subspace of the complete metric space \(\displaystyle{\left(\mathbb{R},\left|\cdot\right|\right)}\), compact

as a closed and bounded subset of the metric space \(\displaystyle{\left(\mathbb{R},\left|\cdot\right|\right)}\) and

separable since \(\displaystyle{\mathbb{Q}\cap \left[0,\dfrac{\pi}{2}\right]\subseteq \left[0,\dfrac{\pi}{2}\right]}\) and

\(\displaystyle{\rm{cl}\left(\mathbb{Q}\cap \left[0,\dfrac{\pi}{2}\right]\right)=\overline{\mathbb{Q}}\cap \left[0,\dfrac{\pi}{2}\right]=\left[0,\dfrac{\pi}{2}\right]}\) or

\(\displaystyle{\rm{cl}\left(\mathbb{Q}\cap \left[0,\dfrac{\pi}{2}\right]\right)=\left[0,\dfrac{\pi}{2}\right]-\rm{Int}\,\left(\left[0,\dfrac{\pi}{2}\right]-\mathbb{Q}\right)=\left[0,\dfrac{\pi}{2}\right]}\) .
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